At 429 K, this reaction has a \( K_c \) value of 0.0407.\[ \text{X(g)} + 2 \text{Y(g)} \rightleftharpoons \text{Z(g)} \]Calculate \( K_p \) at 429 K. Note that the pressure is in units of atmosphere (atm).\[ K_p = \]--- In this exercise, you are asked to calculate the equilibrium constant \( K_p \) at 429 K for a given chemical reaction, using the given \( K_c \) value. The conversion between \( K_c \) and \( K_p \) involves using the equation:\[ K_p = K_c(RT)^{\Delta n} \]Where:- \( R \) is the ideal gas constant (0.0821 L·atm/mol·K).- \( T \) is the temperature in Kelvin.- \( \Delta n \) is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).In this reaction:\[ \Delta n = (\text{moles of Z}) - (\text{moles of X + moles of 2Y}) = 1 - (1 + 2) = -2 \]Use this information to find \( K_p \).

The reaction is, X (g) + 2 Y (g) ⇌ Z (g) For any ideal gas, the…

Answered: At 429 K, this reaction has a Ke value…

At 429 K, this reaction has a Ke value of 0.0407. X(g) + 2 Y(g) = Z(g) Calculate Kp at 429 K. Note that the pressure is in units of atmosphere (atm). Kp =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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